Physics Help Forum What does the following equations prove when equated?

 Special and General Relativity Special and General Relativity Physics Help Forum

 Jun 23rd 2017, 07:58 PM #1 Senior Member     Join Date: Feb 2017 Posts: 205 What does the following equations prove when equated? We know that einsteins formula states E=mc2. Now this is the potential energy. So PE= mgh= mc2. So gh=c2. Does it prove that speed of light is gravitational constant into height of the object?
 Jun 23rd 2017, 08:56 PM #2 Senior Member   Join Date: Apr 2017 Posts: 486 Well it's true to say , in a way, that E=mc2 shows the energy potential of matter ... But this is not the same as the potential energy matter has by having height in a gravitational field .. the two are not the same, so it's not true to say mgh = mc2
 Jun 23rd 2017, 09:07 PM #3 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 550 PE = mgh is an approximate formula for gravitational potential energy. More accurate would be PE = -GmM/r where M is mass of the earth. Most accurate is Einstein's general theory of relativity and gravity. E = mc^2 is more about how mass changes to a "stationary" object if it radiates or loses energy. See this video: and In brief it appears like you're trying to work out the effect of relativistic mass in earths gravitational field but the formulas E=mc^2 and PE = mgh don't seem to me to be in the same league so I'm not sure why someone might do that.
Jun 24th 2017, 11:01 PM   #4
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 Originally Posted by avito009 We know that einsteins formula states E=mc2. Now this is the potential energy. So PE= mgh= mc2. So gh=c2. Does it prove that speed of light is gravitational constant into height of the object?
In general relativity the energy of a body is given by the time component of the time component of the canonical momentum, P[sub]0[/sub]. For a particle of proper mass m[sub]0[/sub] it has the value

P[sub]0[/sub] = m[sub]0[/sub]g[sub]0v[/sub]P[sup]v[/sup] where you sum over v = 0, ..., 3.

The gravitational potential is part of g[sub]0v[/sub]. However its not true that P[sub]0[/sub] = mc[sup]2[/sup]. In fact E = mc[sup]2[/sup] is not an expression which holds in general.

 Jun 25th 2017, 05:05 AM #5 Senior Member     Join Date: Feb 2017 Posts: 205 Can we equate these? If E=mc2 =mgh cant be true then can GMm/r2=mc2 be true?, now m cancels out so we are left with GM/r2=c2. But we know that speed of light is 300000 km/s. If we solve left hand side answer is GM/r2= 9.79811147 m / s2 which doesnt match to speed of light so we cant equate these formulas also. So which formula of potential energy can we equate with mc2?
Jun 25th 2017, 01:09 PM   #6
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 Originally Posted by avito009 If E=mc2 =mgh cant be true then can GMm/r2=mc2 be true?
Don't try to understand physics as though you were simply trying to understand how to understand algebra. Understand the physical concepts first and then let maths follow. I'm a believer in that mathematics is a tool, a "servant" rather than an "oracle". (Maybe everyone may not agree with me on this but it has helped me). In other words the understanding of the physical processes should lead the maths, not the other way around.

Did you see the video above that E=mc^2 is incomplete. Its only half the story. It constitutes only the energy radiated from mass, not the kinetic energy. What you are probably after is Total Energy E = Kinetic energy + potential energy and total energy is always conserved (can't be created or destroyed). So a more useful equation is mv^2/2 = mgh or mv^2/2 = GMm/r (not over r^2, one is a force, the other is an energy potential).

This video animation is the best one I found that explained energy.

 Jun 25th 2017, 02:35 PM #7 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,684 $\displaystyle mc^2$ isn't a potential energy. As PMB and kiwiheretic were saying, $\displaystyle mc^2$ is the energy of a mass that is at rest. If the particle moves you have to account for any momentum. In this respect $\displaystyle E^2 = (mc^2)^2 + (pc)^2$. This reduces to $\displaystyle E = \pm mc^2$ when p = 0. -Dan kiwiheretic likes this. __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.
 Jun 26th 2017, 12:43 AM #8 Senior Member     Join Date: Feb 2017 Posts: 205 My interpretation So E is the hypotenuse and the base is pc. So base cannot be more than hypotenuse. So let us assume E= pc. so after substituting mc2 in E which gives us mc2=pc now substitute mv into p this gives us mc2= mvc after cancelling what remains is v=c. But we know that velocity of an object cant be equal speed of light. so E is not equal to pc. But in that case E=pc in case of light? Am i right? Last edited by avito009; Jun 26th 2017 at 01:24 AM.
Jun 26th 2017, 06:34 PM   #9
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 Originally Posted by avito009 So E is the hypotenuse and the base is pc. So base cannot be more than hypotenuse. So let us assume E= pc. so after substituting mc2 in E which gives us mc2=pc now substitute mv into p this gives us mc2= mvc after cancelling what remains is v=c. But we know that velocity of an object cant be equal speed of light. so E is not equal to pc. But in that case E=pc in case of light? Am i right?
In terms of high speeds near speed of light we need to take into account relativistic mass so $\displaystyle p = mv$ becomes $\displaystyle p = \frac{ m_{0}v}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ where m0 is rest mass. If you are talking about massless particles like photons then it is correct to use E=pc. Otherwise, if the particle has mass, then $\displaystyle E=m c^2$ becomes $\displaystyle E=\frac{m_{0} c^{2}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ and the m0 is again rest mass moving at velocity v. (Note: The E we are talking about here is kinetic, not potential energy.)

Now for low speeds much slower than light we can approximate $\displaystyle E=\frac{m_{0} c^{2}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} \approx \frac{m_{0} v^{2}}{2} + m_{0} c^{2} + \mathcal{O}\left(v^{4}\right)$ using Taylor's series expansion. This shows that for low speeds $\displaystyle \frac{m_{0} v^{2}}{2}$ is a good approximation of kinetic energy.

Hence for low speeds and for ordinary physical conditions like dropping a ball off a cliff then using $\displaystyle E_{kinetic} = \frac{m_{0} v^{2}}{2}$ with $\displaystyle E_{potential} = m g h$ along with conservation of energy principles makes sense. However. trying to use relativistic formulae in such cases is "overkill" and is a pain to calculate minor increases in inertial mass for low velocity when $\displaystyle \frac{m_{0} v^{2}}{2}$ does just as well. So remember when you see E=mc^2 it is not a formula about rest mass (m0) but the inertial mass ($\displaystyle m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ ) which is not a constant for relativistic velocities and is generally not a formula we would use for low velocities when the difference between rest mass and inertial mass are negligible.

Last edited by kiwiheretic; Jun 26th 2017 at 06:40 PM.

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